If a hole was cut straight through the earth to the opposite side and you jumped

Assuming a perfectly spherical earth, equal mass distribution, and no losses of energy (frictionless):

You would accelerate (speed up) until you reach the center of the earth. Once that point is passed, the majority of the earths mass is located towards the direction from which you've been falling. Consequently you decelerate until you reach the other side of the earth. You stop once exactly reaching the other side then fall back towards the other side. You can press escape at anytime. :P Basically you bounce from one side to the other over and over. Now add friction and air resistances, and you don't ever quite reach the other end. Eventually after many oscillations, you come to a halt at the gravitational center.

The real question is this? Why would I jump?GrandJury

Assuming a perfectly spherical earth, equal mass distribution, and no losses of energy (frictionless):

You would accelerate (speed up) until you reach the center of the earth. Once that point is passed, the majority of the earths mass is located towards the direction from which you've been falling. Consequently you decelerate until you reach the other side of the earth. You stop once exactly reaching the other side then fall back towards the other side. You can press escape at anytime. :P Basically you bounce from one side to the other over and over.

coolbeans90

[QUOTE="coolbeans90"]

Assuming a perfectly spherical earth, equal mass distribution, and no losses of energy (frictionless):

You would accelerate (speed up) until you reach the center of the earth. Once that point is passed, the majority of the earths mass is located towards the direction from which you've been falling. Consequently you decelerate until you reach the other side of the earth. You stop once exactly reaching the other side then fall back towards the other side. You can press escape at anytime. :P Basically you bounce from one side to the other over and over.

MrLions

Eh, it'd get boring after a while, but I'd certainly go for it at least once...

uhm..i have no idea...

That's a pretty good question. If you go with just straight physics, then the following would hold true.

Gravity would slow you down just as equally as it sped you up when you jumped down the hole. For every action, there is an opposite and equal reaction. So, let's say that there's nothing to slow you down, or mess with the pull of gravity (a.k.a air resistance). We know things fall at a rate of 9.8m/second squared. The radius of the Earth is 6.378 million meters so you'd make the trip in about 42 minutes (I saw it once on the Discovery channel =D ). You'd basically pop up on the other side of the hole at the same height in which you jumped in and repeat the process unless someone tossed you a rope when you popped out of one of the holes.

Once you throw in air resistance, then you've got to worry about terminal velocities, which will lengthen the trip by days and would probably cause you to bump into the walls of your hole, etc.

That's the quick and short answer for two possible scenarios.

That's a pretty good question. If you go with just straight physics, then the following would hold true.

Gravity would slow you down just as equally as it sped you up when you jumped down the hole. For every action, there is an opposite and equal reaction. So, let's say that there's nothing to slow you down, or mess with the pull of gravity (a.k.a air resistance). We know things fall at a rate of 9.8m/second squared. The radius of the Earth is 6.378 million meters so you'd make the trip in about 42 minutes (I saw it once on the Discovery channel =D ). You'd basically pop up on the other side of the hole at the same height in which you jumped in and repeat the process unless someone tossed you a rope when you popped out of one of the holes.

Once you throw in air resistance, then you've got to worry about terminal velocities, which will lengthen the trip by days and would probably cause you to bump into the walls of your hole, etc.

That's the quick and short answer for two possible scenarios.

joseph_mach

You start at 9.8 m/s^2, but your acceleration gradually drops until hitting 0 m/s^2 at the center of the earth, and then -9.8m/s^2 once arriving at the other side. (assuming frictionless, spherical, and evenly mass distributed earth...)

[QUOTE="joseph_mach"]

That's a pretty good question. If you go with just straight physics, then the following would hold true.

Gravity would slow you down just as equally as it sped you up when you jumped down the hole. For every action, there is an opposite and equal reaction. So, let's say that there's nothing to slow you down, or mess with the pull of gravity (a.k.a air resistance). We know things fall at a rate of 9.8m/second squared. The radius of the Earth is 6.378 million meters so you'd make the trip in about 42 minutes (I saw it once on the Discovery channel =D ). You'd basically pop up on the other side of the hole at the same height in which you jumped in and repeat the process unless someone tossed you a rope when you popped out of one of the holes.

Once you throw in air resistance, then you've got to worry about terminal velocities, which will lengthen the trip by days and would probably cause you to bump into the walls of your hole, etc.

That's the quick and short answer for two possible scenarios.

coolbeans90

You start at 9.8 m/s^2, but your acceleration gradually drops until hitting 0 m/s^2 at the center of the earth, and then -9.8m/s^2 once arriving at the other side. (assuming frictionless, spherical, and evenly mass distributed earth...)

Ah, but in this case, it's gravity that's causing the acceleration. Gravity is the acting force in this equation. F=MA

That's what's fun about physics. I know it sounds strange, but it's the center of the Earth that's pulling us here. So the closer to it we are, the faster we're going to be pulled, again if, as you stated (assuming frictionless, spherical, and evenly mass distributed earth...) are out of the equation.

[QUOTE="coolbeans90"]

[QUOTE="joseph_mach"]

That's a pretty good question. If you go with just straight physics, then the following would hold true.

Gravity would slow you down just as equally as it sped you up when you jumped down the hole. For every action, there is an opposite and equal reaction. So, let's say that there's nothing to slow you down, or mess with the pull of gravity (a.k.a air resistance). We know things fall at a rate of 9.8m/second squared. The radius of the Earth is 6.378 million meters so you'd make the trip in about 42 minutes (I saw it once on the Discovery channel =D ). You'd basically pop up on the other side of the hole at the same height in which you jumped in and repeat the process unless someone tossed you a rope when you popped out of one of the holes.

Once you throw in air resistance, then you've got to worry about terminal velocities, which will lengthen the trip by days and would probably cause you to bump into the walls of your hole, etc.

That's the quick and short answer for two possible scenarios.

joseph_mach

You start at 9.8 m/s^2, but your acceleration gradually drops until hitting 0 m/s^2 at the center of the earth, and then -9.8m/s^2 once arriving at the other side. (assuming frictionless, spherical, and evenly mass distributed earth...)

Ah, but in this case, it's gravity that's causing the acceleration. Gravity is the acting force in this equation. F=MA

That's what's fun about physics. I know it sounds strange, but it's the center of the Earth that's pulling us here. So the closer to it we are, the faster we're going to be pulled, again if, as you stated (assuming frictionless, spherical, and evenly mass distributed earth...) are out of the equation.

No the center of the earth is no pulling us.. Its the fact that when standing on the surface of the planet.. The entire mass of the earth to the opposite side is pulling on you.. The deeper you go the less it will pull you directly below you, because more mass is around you rather then below you.

[QUOTE="joseph_mach"]

[QUOTE="coolbeans90"]

You start at 9.8 m/s^2, but your acceleration gradually drops until hitting 0 m/s^2 at the center of the earth, and then -9.8m/s^2 once arriving at the other side. (assuming frictionless, spherical, and evenly mass distributed earth...)

sSubZerOo

Ah, but in this case, it's gravity that's causing the acceleration. Gravity is the acting force in this equation. F=MA

That's what's fun about physics. I know it sounds strange, but it's the center of the Earth that's pulling us here. So the closer to it we are, the faster we're going to be pulled, again if, as you stated (assuming frictionless, spherical, and evenly mass distributed earth...) are out of the equation.

No the center of the earth is no pulling us.. Its the fact that when standing on the surface of the planet.. The entire mass of the earth to the opposite side is pulling on you.. The deeper you go the less it will pull you directly below you, because more mass is around you rather then below you.

Here's a great page that explains the "physics" part of it. Again, I just wrote down the short of it, taking more than half of the variables out, which is why I said two possible scenarios in my original post.

From PhysicsCentral

[QUOTE="joseph_mach"]

[QUOTE="coolbeans90"]

You start at 9.8 m/s^2, but your acceleration gradually drops until hitting 0 m/s^2 at the center of the earth, and then -9.8m/s^2 once arriving at the other side. (assuming frictionless, spherical, and evenly mass distributed earth...)

sSubZerOo

Ah, but in this case, it's gravity that's causing the acceleration. Gravity is the acting force in this equation. F=MA

That's what's fun about physics. I know it sounds strange, but it's the center of the Earth that's pulling us here. So the closer to it we are, the faster we're going to be pulled, again if, as you stated (assuming frictionless, spherical, and evenly mass distributed earth...) are out of the equation.

No the center of the earth is no pulling us.. Its the fact that when standing on the surface of the planet.. The entire mass of the earth to the opposite side is pulling on you.. The deeper you go the less it will pull you directly below you, because more mass is around you rather then below you.

Yup. It's mass that causes gravitational pull.

[QUOTE="sSubZerOo"]

[QUOTE="joseph_mach"]

Ah, but in this case, it's gravity that's causing the acceleration. Gravity is the acting force in this equation. F=MA

That's what's fun about physics. I know it sounds strange, but it's the center of the Earth that's pulling us here. So the closer to it we are, the faster we're going to be pulled, again if, as you stated (assuming frictionless, spherical, and evenly mass distributed earth...) are out of the equation.

joseph_mach

No the center of the earth is no pulling us.. Its the fact that when standing on the surface of the planet.. The entire mass of the earth to the opposite side is pulling on you.. The deeper you go the less it will pull you directly below you, because more mass is around you rather then below you.

Here's a great page that explains the "physics" part of it. Again, I just wrote down the short of it, taking more than half of the variables out, which is why I said two possible scenarios in my original post.

From PhysicsCentral

It's using the universal gravitational constant. Not the acceleration due to gravity on earth. HUGE difference. They are related, but not the same. The thing about that is the mass of the gravitational pull changes.

Anyways, quotes from your source.

"But when you start falling, things start changing. As you fall there is less and less mass between you and the center so there is less and less to pull you down."

"The acceleration due to gravity depends on the mass that's pulling so once you being to fall, your acceleration gets smaller." See, it starts at 9.8 m/s^2, but reduces to zero at the middle, leaving the velocity (decreasing at that point) to bring you the rest of the way.

[QUOTE="joseph_mach"]

[QUOTE="sSubZerOo"]

No the center of the earth is no pulling us.. Its the fact that when standing on the surface of the planet.. The entire mass of the earth to the opposite side is pulling on you.. The deeper you go the less it will pull you directly below you, because more mass is around you rather then below you.

coolbeans90

Here's a great page that explains the "physics" part of it. Again, I just wrote down the short of it, taking more than half of the variables out, which is why I said two possible scenarios in my original post.

From PhysicsCentral

It's using the universal gravitational constant. Not the acceleration due to gravity on earth. HUGE difference. They are related, but not the same. The thing about that is the mass of the gravitational pull changes.

Anyways, quotes from your source.

"But when you start falling, things start changing. As you fall there is less and less mass between you and the center so there is less and less to pull you down."

"The acceleration due to gravity depends on the mass that's pulling so once you being to fall, your acceleration gets smaller." See, it starts at 9.8 m/s^2, but reduces to zero at the middle, leaving the velocity (decreasing at that point) to bring you the rest of the way.

Which is exactly what I'm saying, while not saying it in a thousand words my friend. Your velocity will carry you through, as the opposite force now pulls on you, slowing you down, and you pop up on the other side of the world (without air to interfere). I just gave the short answer using 1 scenario, using, shall I say "abbreviated" physics.

[QUOTE="coolbeans90"]

[QUOTE="joseph_mach"]

Here's a great page that explains the "physics" part of it. Again, I just wrote down the short of it, taking more than half of the variables out, which is why I said two possible scenarios in my original post.

From PhysicsCentral

joseph_mach

It's using the universal gravitational constant. Not the acceleration due to gravity on earth. HUGE difference. They are related, but not the same. The thing about that is the mass of the gravitational pull changes.

Anyways, quotes from your source.

"But when you start falling, things start changing. As you fall there is less and less mass between you and the center so there is less and less to pull you down."

"The acceleration due to gravity depends on the mass that's pulling so once you being to fall, your acceleration gets smaller." See, it starts at 9.8 m/s^2, but reduces to zero at the middle, leaving the velocity (decreasing at that point) to bring you the rest of the way.

Which is exactly what I'm saying, while not saying it in a thousand words my friend. Your velocity will carry you through, as the opposite force now pulls on you, slowing you down, and you pop up on the other side of the world (without air to interfere). I just gave the short answer using 1 scenario, using, shall I say "abbreviated" physics.

Understood. The gravitational acceleration of 9.8 m/s^2 still only applies at the surfaces. It's not constant. Just had to point that out.

[QUOTE="markop2003"][QUOTE="argetlam00"]

At the center of the earth, assuming there is gravity, you will simply be tossed up and down. As you pass the core, you will be pulled upwards back towards the core. Then as you pass the core again, you will deaccelerate and then head downwards towards the core again. This pattern will constantly continue. The direction of the force of gravity will be constantly changing, therefore your position-time graph would probably look similar to that of a sine graph.

argetlam00

ACtually there is a force of gravity buddy. It stems from the basic equation of newton second law F=ma, changed to F=mg....This is commonly reffered to the force of gravity.

there's a mother****ing hole in the planet.Theokhoth

I'm sure you'd just pop up next to some very sterotypical chinese people and everything would be cool. Atleast that's what I'm led to believe.

[QUOTE="Perd1t1on"][QUOTE="xaos"] Except that you'd fall a bit less on each trip through due to friction lossesmagnax1

No, but it'd be a constant force, so the friction wouldn't slow you down. Its like being dragged on the ground, the ground slows you but as long as you're being pulled you'd keep going.

[QUOTE="argetlam00"]

[QUOTE="markop2003"] Gravity is an acceleration not a force /smartass Oh and you'ld be correct if it was a vacuum but i think the TC meant that you'ld somehow survive considering he was talking about starving to death, so i assume there must be some air.Dark__Link

ACtually there is a force of gravity buddy. It stems from the basic equation of newton second law F=ma, changed to F=mg....This is commonly reffered to the force of gravity.

Aren't there 4 "known forces" in physics? Electromagnetic, Weak and Strong Nuclear, and Gravity? Maybe it's a bit later than I thought it was and I'm pretty sleepy, but I'm pretty sure gravity is at least "listed" as one of the 4 known forces. I mean I understand the relationship between gravity and acceleration, but I think it's listed as a force right?

[QUOTE="magnax1"]

[QUOTE="Perd1t1on"] given that there is air resistance. wouldn't it be a vacuum in there?xaos

No, but it'd be a constant force, so the friction wouldn't slow you down. Its like being dragged on the ground, the ground slows you but as long as you're being pulled you'd keep going.

True story.

[QUOTE="argetlam00"]

[QUOTE="markop2003"] Gravity is an acceleration not a force /smartass Oh and you'ld be correct if it was a vacuum but i think the TC meant that you'ld somehow survive considering he was talking about starving to death, so i assume there must be some air.Dark__Link

ACtually there is a force of gravity buddy. It stems from the basic equation of newton second law F=ma, changed to F=mg....This is commonly reffered to the force of gravity.

And just for the record, I have to say this is one of the better threads I've seen in a long time. :) At least in my humble opinion. It's far more interesting than the normal "soda or tea" threads that have been "all the rage" lately.

Kudos TC!

Because of air resistance you would end up stuck in the center of the earth.

[QUOTE="Dark__Link"][QUOTE="argetlam00"]

ACtually there is a force of gravity buddy. It stems from the basic equation of newton second law F=ma, changed to F=mg....This is commonly reffered to the force of gravity.

xaos

well I guess you would fall to the center and as you reach the center you'd slow down to a stand still. Then at that point youd have to climb up against gravity. I think it would be a little easier to climb the first few km but it would get harder gradually. Then you'd hit the crust and be in the ocean. If you hadn't already died from the incredible pressure, in this hypothetical situation you would need to swim to the top and you would be on the other side of the world.

And just for the record, I have to say this is one of the better threads I've seen in a long time. :) At least in my humble opinion. It's far more interesting than the normal "soda or tea" threads that have been "all the rage" lately.

Kudos TC!

joseph_mach

[QUOTE="xaos"][QUOTE="Dark__Link"] Uh... no. Gravity is not a force. You do know what the "a" and "g" stand for in those equations, right? Oh wait no, you must not...Dark__Link

[QUOTE="Dark__Link"]

[QUOTE="xaos"] Gravity is a force, but argetlam00 is inferring the wrong meaning from the acceleration formula. It just indicates that amount of force required to given a particular mass a particular acceleration; the only classical formula relating directly to gravity is the law of universal gravitation.xaos

We would all die, because there's a mother****ing hole in the planet.Theokhoth

For the sake of simplicity lets say their is no atmosphere, no planetary rotation and equal density. Your speed would increase until you reached the middle of the planet at which point the gravity of the planet would be pulling at you equally from all directions and your acceleration would be zero. As you fell further and further towards the other side of the planet, your speed would continue to decrease until it would be approximately nil at the time you popped up on the other side thus conserving momentum.

At least that's how I understand it, but I'm no physics major.

It's called the Gravity Train.

You'd collide with a dumb *** deer coming from the opposite direction...

k, everybody...I have found the answer online and will summarize:

-This assumes there is uniform density, no air friction, and no high temperatures

-As stated before you will oscillate back and forth indefinitely, the full trip being t=84.5minutes

This is the main equation, mass (m) cancels out.

-The velocity at the center is 7900m/s

If there was air friction, you wouldn't travel as fast. I could not find info for that. This equation is my guess as to why you would move less each time:

v^2=v.^2-2a(x-x.)

Since v=less every time (and v. is 0 at both ends, while a=g remains the same), the displacement of the distance gets smaller and smaller each time too. And so you eventually stay put in the center.

As a side note, given you can breathe, you'd probably throw up your organs from motion sickness and die anywayfalling at17,700mi/hour (7900m/s).

source: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

I made a picture for the scenario in a thread about a sinkhole a month or two ago!

It's scientific, and to-scale.

We would all die, because there's a mother****ing hole in the planet.Theokhoth

I agree with this :lol:

If you jumped you'd oscillate between both sides of the earth. The gravitation force you feel is proportional to the amount of spherical earth below you. So the gravity gets less and less as you get closer to the core. It approaches zero as you reach the center and then starts pulling towards the other way. Theoretically, you should come out the other side at the exact same height above the surface as you jumped from.

(assuming the earth did not rotate and there was no atmosphere...)

Did some quick physics and calculated the your acceleration would be:

4GD(pi)r/3

G being the gravity constant, D being earth's average density and r being the distance from the core of the earth you are.

The differential equation for your position is give as follows:

mx'' + 4GD(pi)mr/3 = 0

And I think that works out to

x = cSin(root(4GDpi/3)t)

(yeah, I'm bored)